Suppose $G$ is abelian and the order of $G$ is $h$ Let $g_1$ $g_2,\ldots, g_h$ be the elements of $G$ and for $i=1,\ldots, h$ let $a_i$ be the order of $g_i$

Consider the direct sum $$ H = \bigoplus_{i=1}^h \mathbb{Z}/a_i\mathbb{Z}. $$ The order of $H$ is obviously $a_1a_2\cdots a_h$ We can define a group homomorphism $\theta$ from $H$ to $G$ by $$ (x_1,\ldots,x_h) \mapsto g_1^{x_1}\cdots g_h^{x_h}. $$ $\theta$ is certainly surjective. So $|H| = |G|\cdot|\ker(\theta)|$ Since $p$ is a prime factor of $G$ $p$ divides |H|, and therefore must divide one of the $a_i$ s, say $a_1$ Then $g_1^{a_1/p}$ is an element of order $p$